Question

pls keep ans with solution then l will mark u as brainlist​

Answer 1

Answer:

hey mate 2nd is the correct answer

Explanation:

PlZ mark me as brainlist

Answer 2

Answer:

Given that

Exposure limit =35 ppm=35 mg/L

Density of air =1.3 g/L

Volume of air =1.0 g/L

Mass of air =1.3 gL  ^−1

×1.0L=1.3 g

Mass of CO in 1.3 g air =1.3×35 mg/L

=45.5 mg/L

=45.5×10  ^−3

g/L

=45.5×10  ^−6

g/mL

=4.55×10  ^−5  g in per mL air

pls mark me as a  Brainliest ;)